# Find the extremal to the functional and discuss whether they provide a max/min

#### dtwazere

I am having a hard time getting my head around Functionals and Calculus of Variations,

My question is: Given a functional and using the Euler-Lagrange equation to find an extremal, how do we show that the extremal provides a min/max (if it does)

The question I am working on is

$$\displaystyle J(y) = \int_{0}^{1} ((y')^2 -y)dx$$ with $$\displaystyle y(0)=0, y(1)=1$$

I found the extremal to be: $$\displaystyle y(x) = \frac{-1}{4}x^2 +\frac{5}{4}x$$ which I am told is a minimum to the functional problem.

However I am unsure on what is sufficient to show this, in the notes I have it is shown that:

$$\displaystyle J(y+f) = J(y) + \int_{0}^{1}(f')^2dx \geq J(y)$$ where f is continuously differentiable on the interval 0,1 with $$\displaystyle y(0)=y(1)=0$$

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#### GJA

Hi dtwazere.

Calculus of Variations can seem a little daunting, however hopefully a few comments will make things seem a little less scary.

1) You can think of the term "functional" in this case as something that sends functions to real numbers. The definition of a functional is actually a little more technical than that (it is a map from a vector space to its underlying scalar field - see Functional (mathematics) - Wikipedia, the free encyclopedia) - in your case the vector space is the space of continuously differentiable functions with $$\displaystyle y(0)=0$$ and $$\displaystyle y(1)=1$$ and the "field" is the real numbers. However, thinking of a functional as something whose domain is a certain set of functions (in your case continuously differentiable functions with $$\displaystyle y(0)=0$$ and $$\displaystyle y(1)=1$$) and whose range is the real numbers is good enough for now.

2) You have actually done most of the hard work already by solving the Euler-Lagrange equation to find a possible minimizer. I will denote your possible minimizer by $$\displaystyle y_{min};$$ i.e. $$\displaystyle y_{min}(x)=-\frac{1}{4}x^{2}+\frac{5}{4}x.$$ Note that my notation is a little dangerous because we haven't actually proved that $$\displaystyle y_{min}$$ is actually a minimizer of $$\displaystyle J$$ yet. The last thing we need to note is that the condition $$\displaystyle f(0)=0=f(1)$$ was left out in the original post where you mentioned $$\displaystyle J(y+f).$$ The condition $$\displaystyle f(0)=1=f(1)$$ is essential; $$\displaystyle y_{min}+f$$ represents a small pertubation of $$\displaystyle y_{min}$$, however we don't perturb the endpoints of $$\displaystyle y_{min}$$, which is where the condition $$\displaystyle f(0)=1=f(1)$$ comes from.

Now we compute

$$\displaystyle J(y_{min}+f)=\int_{0}^{1}[(y_{min}+f)']^{2}-(y_{min}+f)dx$$

expanding this we get

$$\displaystyle J(y_{min}+f)=\int_{0}^{1}(y_{min}')^{2}-y_{min}dx+\int_{0}^{1}2y_{min}'f'dx-\int_{0}^{1}fdx+\int_{0}^{1}(f')^{2}dx$$

Which becomes

$$\displaystyle J(y_{min}+f)=J(y_{min})+\int_{0}^{1}2y_{min}'f'dx-\int_{0}^{1}fdx+\int_{0}^{1}(f')^{2}dx$$

If we integrate by parts on the second term on the RHS we obtain
$$\displaystyle 2\int_{0}^{1}y_{min}'f'dx=2(y'f |_{0}^{1} -\int_{0}^{1}y_{min}''fdx)$$

Now use $$\displaystyle f(0)=0=f(1)$$ and $$\displaystyle y_{min}''=-\frac{1}{2}$$ on the previous line to get

$$\displaystyle 2\int_{0}^{1}y_{min}'f'dx=\int_{0}^{1}fdx$$

Using the last line in $$\displaystyle J(y_{min}+f)$$ we have

$$\displaystyle J(y_{min}+f)=J(y_{min})+\int_{0}^{1}fdx-\int_{0}^{1}fdx+\int_{0}^{1}(f')^{2}dx=J(y_{min})+\int_{0}^{1}(f')^{2}dx\geq J(y_{min})$$

This proves that $$\displaystyle y_{min}$$ really is a minimizer of $$\displaystyle J$$ for the class of functions that you're considering.

Does this help? Let me know if anything is unclear.

Good luck!

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#### dtwazere

This really helps, I think I understand most of what you put, thanks a lot!